1/π = 12 Σ c (16254n+789)/(9603)n+1/2
1/π = 12 Σ c (261702n+10177)/(52803)n+1/2
1/π = 12 Σ c (545140134n+13591409)/(6403203)n+1/2
where the sum Σ goes from n = 0 to ∞. Beautiful, aren’t they? The modular function responsible for this is the j-function which has the q-series expansion,
j(q) = 1/q + 744 + 196884q + 21493760q2 + 864299970q3 + ...
which explains why for these discriminants, the number eπ√d has an excess close to the integer "744". This is sequence A000521 of the OEIS and, other than the constant term, is the McKay-Thompson series class 1A of the Monster simple group. These pi formulas were discovered by the Chudnovsky brothers inspired by Ramunujan’s formulas given in the next section (which uses d with class number 2). For more details, see “Pi formulas, Ramanujan, and the Baby Monster Group”, by this author (File 07).
II. Class number h(-d) = 2, where d = 4m. There are exactly 18 negative fundamental discriminants d with class number h(-d) = 2, 7 of which are even, namely d = 4m, for m = {5, 13, 37} and m = {6, 10, 22, 58}. As m increases, the expression eπ√m gets intriguingly close to an integer with a fixed excess "104",
eπ√5 ≈ 322 + 100
eπ√13 ≈ 2882 + 103.9…
eπ√37 ≈ 141122 + 103.9999…
eπ√6 ≈ 482 - 106
eπ√10 ≈ 1442 - 104.2…
eπ√22 ≈ 15842 - 104.001…
eπ√58 ≈ 1568162 - 104.0000001…
Just like for d with class number h(-d) = 1, there are also nice Diophantine relationships between these numbers,
5 (44 + 322) = 42 *202
13 (44 + 2882) = 42 *2602
37 (44 + 141122) = 42 *214602
6 (-44 + 482) = 82 *3*82
10 (-44 + 1442) = 82 *2*402
22 (-44 + 15842) = 82 *11*2802
58 (-44 + 1568162) = 82 *2*1055602
So the sums for the first set are perfect squares, while the second set are almost-squares. The blue and red numbers again appear in pi formulas courtesy of, who else, but Ramanujan. Let r = (4n)! / (n!4), then,
1/π = 4 Σ r (-1)n (20n+3)/(322)n+1/2
1/π = 4 Σ r (-1)n (260n+23)/(2882)n+1/2
1/π = 4 Σ r (-1)n (21460n+1123)/(141122)n+1/2
1/π = 8√3 Σ r (8n+1)/(482)n+1/2
1/π = 8√2 Σ r (40n+4)/(1442)n+1/2
1/π = 8√11 Σ r (280n+19)/(15842)n+1/2
1/π = 8√2 Σ r (105560n+4412)/(1568162)n+1/2
where n = 0 to ∞. The modular function, call it r(q), that is responsible for this has the q-series expansion,
r(q) = 1/q + 104 + 4372q + 96256q2 + 1240002q3 + …
and explains why the number eπ√m has an excess close to the integer "104". This is sequence A007267 and, excepting the first term, is the McKay-Thompson series of Class 2A for the Monster group.
III. Class number h(-d) = 2, where d = 3m. There are exactly 5 discriminants of this form, namely d = 3m, for m = {5, 8, 17, 41, 89}. This time, the expression eπ√(m/3) gets close to an integer with a fixed excess "42", though I will focus only on the three highest m,
eπ√(17/3) ≈ 123 + 41.6…
eπ√(41/3) ≈ 483 + 41.99…
eπ√(89/3) ≈ 3003 + 41.9999…
and three non-fundamental odd m = {9, 25, 49},
eπ√(9/3) ≈ 3*43 + 38.8…
eπ√(25/3) ≈ 5*123 + 41.91…
eπ√(49/3) ≈ 7*363 + 41.998…
Since these non-fundamental m are the squares of the first 3 odd primes, the approximations look nice, being the product of that prime and a cube. As usual, there are nice Diophantine relationships,
3*17 (2233 + 123) = 62 *512
3*41 (2233 + 483) = 62 *6152
3*89 (2233 + 3003) = 62 *141512
3 (2233 + 3*43) = 62 *52
3 (2233 + 5*123) = 62 *272
3 (2233 + 7*363) = 62 *1652
so the sums are perfect squares. Define r2 = (2233)n (1/2)n(1/3)n(2/3)n / (n!3) where (a)n is the rising factorial, or Pochhammer symbol, such that (a)n = (a)(a+1)(a+2)…(a+n-1). Equivalently, let r2 = (2n)! (3n)!/ (n!5), then,
1/π = 1/(12√3) Σ r2 (-1)n (51n+7)/(123)n
1/π = 1/(96√3) Σ r2 (-1)n (615n+53)/(483)n
1/π = 1/(1500√3) Σ r2 (-1)n (14151n+827)/(3003)n
1/π = (1/4)√3 Σ r2 (-1)n (5n+1)/(3*43)n
1/π = (1/36)√15 Σ r2 (-1)n (27n+3)/(5*123)n
1/π = (1/108)√7 Σ r2 (-1)n (165n+13)/(7*363)n
where n = 0 to ∞. These have been derived by this author (On Ramanujan's Other Pi Formulas, file 08), though Chan and Liaw gave a different form for the first one in their paper, "Cubic Modular Equations and New Ramanujan-Type Series for 1/π" (2000). The modular function responsible, call it h(q), has q-series expansion,
h(q) = 1/q + 42 + 783q + 8672q2 + 65367q3 + …
This is sequence A030197 and, perhaps not surprisingly, is the McKay-Thompson series of class 3A for the Monster group. To summarize, there is this deep relationship between transcendental numbers of the form eπ√d, the modular functions j(q), r(q), h(q), pi formulas, and the Monster group. (End update.)
(Update, 6/15/10): It is also well known that,
33+43+53 = 63
It seems sums of n 5th powers equal to a 5th power also like the number 6, or specifically, 12. For n = 4,5,6, the smallest solns have sums that are ALL divisible by 12, namely,
4 terms: 275 + 845 + 1105 + 1335 = 1210
5 terms: 195 + 435 + 465 + 475 + 675 = (6*12)5
6 terms: 45 + 55 + 65 + 75 + 95 + 115 = 125
This numerical curiosity is perhaps another instance of Richard Guy’s Strong Law of Small Numbers.
(Update, 6/14/10): Duncan Moore observed that the three known solns to,
x15+x25+x35+x45+x55 = 0 (eq.1)
obey curious congruence properties between its terms (just like this author’s observations about certain 4th power Diophantine equations. Refer to 3/28/10 update.) Given.
1. [27, 84, 110, 133, -144]5 = 0
2. [220, -5027, -6237, -14068, 14132]5 = 0
3. [55, 3183, 28969, 85282, -85359]5 = 0
Scher and Seidl, discoverer of [2], proved that eq.1 obeys the constraint that the sum of either one or two pairs of terms is divisible by 5. Moore observed that the sum of one or two pairs is also divisible by 25, namely,
1. 27 + 133 = 25*5
2. 5027 + 6237 = 11*210, and -14068 + 14132 = 26
3. 55 + 28969 = 907*25, and -3183 + 85359 = 327*28
Whether this is as universally valid as the Scher-Seidl congruence is not known.
(Update, 6/13/10): Alain Verghote pointed out that A. Martin’s four-square identity,
(4pr+s)2 + (4pr-s)2 + (4qr+s)2 + (4qr-s)2 = 4(s+2r2)2, if s = 2p2+2q2-r2
has a second similar soln as,
(4pr+s)2 + (4pr-s)2 + (4qr+s)2 + (4qr-s)2 = 4(s+4r2)2, if s = p2+q2-2r2
Verghote also found the nice 4th power identity,
8(ac-bd)4 + 8(ad-bc)4 = ((a-b)(c+d))4 + 6((a2-b2)(c2-d2))2 + ((a+b)(c-d))4
(Update, 6/12/10): Seiji Tomita gave a new method to solve x3+y3+z3 = mt3 for an infinite number of constant m that is not m ≡ 4,5 (mod 9). While Ryley’s Identity is for any m, Tomita’s method involves polynomials of smaller degree (cubic). Examples are,
(2n3+6n-8)3 + (-2n3-6n-8)3 + (6n2+10)3 = 3(2n2-2)3
(n3+9n2+39n+91)3 + (-n3-9n2-39n-35)3 + (6n2+36n+70)3 = 6(4n2+24n+56)3
and so on. See http://www3.alpha-net.ne.jp/users/fermat/dioph74e.html for more details. He also found,
(-3n3+1)3 + (3n3+2)3 + (3n)3 = (9n3+3)2
similar to Bouniakowsky’s,
(n3+1)3 + (-n3+2)3 + (3n)3 = (3n3+3)2
(Update, 6/11/10): Wroblewski found three solns to the septic 3-way chain,
x1k+x2k+x3k-x4k-x5k-x6k = y1k+y2k+y3k-y4k-y5k-y6k = z1k+z2k+z3k-z4k-z5k-z6k
for k = 1,2,3,5,7. They also obey,
x1k+x2k+x3k = y1k+y2k+y3k = z1k+z2k+z3k
x4k+x5k+x6k = y4k+y5k+y6k = z4k+z5k+z6k
for k = 1,3. Two solns are:
1: {x,y,z} = {a,b,c}
{a1, a2, a3, a4, a5, a6} = {723, -547, -813, -855, 279, 821}
{b1, b2, b3, b4, b5, b6} = {-697, -263, 323, 421, 395, -571}
{c1, c2, c3, c4, c5, c6} = {-703, -317, 383, 509, 341, -605}
2: {x,y,z} = {d,e,f}
{d1, d2, d3, d4, d5, d6} = {-827, -757, 947, 1045, 209, -1009}
{e1, e2, e3, e4, e5, e6} = {47, -157, -527, -205, -59, 509}
{f1, f2, f3, f4, f5, f6} = {227, -417, -447, -349, 45, 549}
Piezas observed that these two solns obey the additional cubic 6-way chain,
a1k+a2k+a3k-a4k-a5k-a6k = b1k+b2k+b3k-b4k-b5k-b6k = c1k+c2k+c3k-c4k-c5k-c6k = d1k+d2k+d3k-d4k-d5k-d6k = e1k+e2k+e3k-e4k-e5k-e6k = f1k+f2k+f3k-f4k-f5k-f6k
for k = 1,3. It is unknown if an identity is behind this septic 3-way, in contrast to the octic 3-way chain where an identity has been found.
(Update, June 2010): Okay, so this is not about math, but I thought I'd share this clip from one of my favorite films from childhood. It's Rankin Bass' Flight of Dragons (1982), an echanting tale about the battle between science and magic. The beautiful opening song is by Don McLean, the singer of that classic song, "American Pie". Sigh, they don't make animated movies like this anymore. Maybe they can make a live-action remake; if it can be done for Lord of the Rings, it certainly can be done for this one.
1/π = 3 Σ c (63n+8)/(153)n+1/2
1/π = 4 Σ c (154n+15)/(323)n+1/2
1/π = 12 Σ c (342n+25)/(963)n+1/2
19 (123 + 963) = 122 *3422
43 (123 + 9603) = 122 *162542
67 (123 + 52803) = 122 *2617022
163 (123 + 6403203) = 122 *5451401342
So the sums are perfect squares. It can also be seen that the equation,
x(123 + Round[(eπ√x - 744)1/3]3) = y2
where Mathematica's "Round[n] function" rounds off n to the nearest integer has integer solns only for x = {7, 11, 19, 43, 67, 163} for x < 10,000. And where else do the blue and red numbers appear? Of all places, in formulas for π! Let c = (-1)n(6n)! / ((n!)3(3n)!), then,
7 (123 + 153) = 32 *632
11 (123 + 323) = 42 *1542
4. Haldeman-Ramanujan quintuples:
22 + 22 + 42 + 32 + 42 = 72
24 + 24 + 44 + 34 + 44 = 54
A Haldeman-Ramanujan quintuple {a,b,c,d,e} solves the system,
a2 + b2 + c2 + d2 + e2 = x2
a4 + b4 + c4 + d4 + e4 = y4
in the integers. There are many small solns, and it can be given an infinite number of simple identities, one of which is,
Pk := (2x2+12xy-6y2)k + (2x2-12xy-6y2)k + (4x2-12y2)k + (3x2+9y2)k + (4x2+12y2)k
where,
P2 = 72(x2+3y2)2
P4 = 54(x2+3y2)4
Haldeman-Ramanujan quintuples with the constraint a+b = c (in blue) can then be used to generate similar polynomial identities (see Form 21). Since the integers include zero, there are in fact solns when a = 0, though the only one known is,
02 + 159352 + 270222 + 579102 + 592602 = 885972
02 + 159354 + 270224 + 579104 + 592604 = 701214
but it is unknown if this special case has an infinite number of solns.
(Update, 6/22/10): The system,
x1x2x3 = y1y2y3
x13+x23+x33 = y13+y23+y33
has several solns:
1. Gerardin
(a2p)3 + (b2q)3 + (abr)3 = (a2q)3 + (-b2r)3 + (-abp)3
where {p,q,r} = {a3-2b3, a3+b3, 2a3-b3}.
2. Choudhry gave a 3-parameter version,
(ap)3 + (bq)3 + (cr)3 = (aq)3 + (-br)3 + (-cp)3
where {p,q,r} = {a3-2b3-c3, a3+b3+2c3, 2a3-b3+c3}.
3. Piezas
(ap)3 + (bq)3 + (cr)3 = (ar)3 + (bp)3 + (cq)3
where {p,q,r} = {ab-c2, -a2+bc, -b2+ac}.
The last identity is also valid for x1+x2+x3 = y1+y2+y3 = 0, as well as,
9x1x2x3 (x16+x26+x36-y16-y26-y36) = 2(x19+x29+x39-y19-y29-y39)
(Update, 6/21/10): For the Mar 22 update, A. Verghote gave the nice identity,
(x+y)(x+2y)(x+3y)(x+4y) + y4 = (x2+5xy+5y2)2 (eq.1)
This has connections to Brocard’s Problem which asks to find values of n for which n!+1 = m2, the smallest of which is n = 4 giving (1)(2)(3)(4) + 1 = 52. Eq.1 can be generalized, with another soln as,
(ax+y)(bx+2y)(cx+3y)(dx+4y) + y4 = (15x2+19xy+5y2)2
where {a,b,c,d} = {1, 3, 5, 15}, so abcd = 152, and an infinite more. This can be generalized for higher n by using the broader n!+v2 = m2 to increase the number of factors. For n = 6, I found,
(ax+y)(bx+2y)(cx+3y)(dx+4y)(ex+5y)(fx+6y) + 9y6 = (12x3+54x2y+72xy2+27y3)2
where {a,b,c,d,e,f} = {2, 2, 3, 2, 2, 3}, and other equivalent forms where abcdef = 122. Is there a soln to the next level n = 8, where n!+92 = 2012, so,
(ax+y)(bx+2y)(cx+3y)…(hx+8y) + 81y8 = (px4+qx3y+rx2y2+sxy3+201y4)2
and the coefficients are integers? (A simple Mathematica code can show there are none if the coefficients are positive integers ≤ 5, but my code takes too long if the range is extended.)
(Update, 6/20/10): Wroblewski has a database of primitive solns (when all terms do not have a common factor) to the equation,
x14+x24+x34+x44+x54 = y14
separated into whether the sum y1 is integrally divisible by 5, or not. The cases x1 = 0, or x1 = x2 = 0, are solvable. If we consider the case where x1 almost vanishes, then for y1 not div by 5, out of more than 13000 solns with sum y1 < 10000, only four examples were found with x1 = 1,
14 + 1504 + 33404 + 61304 + 63504 = 74994
14 + 25204 + 32504 + 50504 + 69704 = 74994
14 + 9204 + 31204 + 54104 + 88704 = 91934
14 + 14104 + 34904 + 60204 + 86804 = 91934
Why it comes in pairs in unknown. It is also unknown if x1 = 0 with x2 = 1 is possible, though the closest is a single example for x2 = 2,
24 + 150454 + 268704 + 340904 = 372394
also found in Wroblewski’s database for (4.1.4) with sum y1 < 222000. A higher search radius might eventually lead to a soln. Other interesting results are,
10504 + 14004 + 14304 + 16654 + 15624 = 328
7354 + 32204 + 37804 + 41604 + 59364 = 332
Note that,
37-1562 = 54
38-5936 = 54
Also,
28334 + 37104 + 72704 + 117204 = 2312
6084 + 24504 + 115304 + 348654 = 1878
16304 + 210194 + 223404 + 339404 = 1918
which are the only three (4.1.4) with such sums out of the about 1000 known. For 6th powers, no known (6.1.7) have similar sums. However, the eqn,
x16+x26+x36+x46+x56 = x66+x76
was the subject of a recently finished distributed computing project which found 181 primitive solns with x7 < 76 using a total of 740 MHz years! It is conjectured that either x5 = 0, or x7 = 0 is solvable, though none have yet been found. For the latter, the smallest known for the moment is x7 = 125 = 53,
2146 + 29686 + 69516 + 200466 + 234576 = 247816 + 518
found by Larry Hays, hence is a relatively close approximation to five 6th powers equal to a 6th power (with a very neat excess).
(Update, 6/15/10 & 6/16/10)
I. Class number h(-d) = 1.
II. Class number h(-d) = 2, where d = 4m.
III. Class number h(-d) = 2, where d = 3m.
I. Class number h(-d) = 1. It is quite well-known that,
eπ√7 ≈ 153 + 697
eπ√11 ≈ 323 + 738
eπ√19 ≈ 963 + 743
eπ√43 ≈ 9603 + 743.999…
eπ√67 ≈ 52803 + 743.99999…
eπ√163 ≈ 6403203 + 743.99999999999…
The discriminants d under the square root are the six highest of the nine Heegner numbers. As I pointed out in a 2008 sci.math.research post, it is not so commonly known there is another internal structure,
eπ√7 ≈ 33(32-22)3 + 697
eπ√11 ≈ 43(32-1)3 + 738
eπ√19 ≈ 123(32-1)3 + 743
eπ√43 ≈ 123(92-1)3 + 744
eπ√67 ≈ 123(212-1)3 + 744
eπ√163 ≈ 123(2312-1)3 + 744
The reason for the squares is due to certain Eisenstein series. But there are other nice Diophantine relationships between these numbers,
10 + 13 + 14 + 44 = 92
102 + 132 + 142 + 442 = 492
103 + 133 + 143 + 443 = 453
(Can anyone find another small solution with no common square factor?) (Update, 6/24/10): James Dow Allen found the three smallest {a,b,c,d} < 1000:
{10, 13, 14, 44}
{54, 109, 202, 260}
{102, 130, 234, 318}
(Update, 6/26/10): Seiji Tomita found the next two with terms < 5000:
{198, 630, 1594, 1674}
{570, 742, 1094, 1690}
Note that these five smallest Martin quadruples (with no common square factor) have very interesting sums,
10 + 13 + 14 + 44 = 34
54 + 109 + 202 + 260 = 54
102 + 130 + 234 + 318 = 2472
198 + 630 + 1594 + 1674 = 212
570 + 742 + 1094 + 1690 = 212
The 6th one is still unknown though. There are also an infinite number of distinct Martin quadruples unscaled by a square factor, as each can be used in an elliptic curve to generate more. For ex, using the first, we have the polynomial,
Pk := (10v2-v+88)k + (13v2+68v+28)k + (14v2-68v+26)k + (44v2+v+20)k
Again, P3 is already a perfect cube for any v, but P2 is a quartic polynomial in v which must be made a square. An initial point is v = 165/94 which yields, after removing unwanted factors, the new soln,
{a,b,c,d} = {-443794m, 1656013m, 1390130m, 1034308m}
where m is to be chosen as in Martin triples. As usual, from this initial point v, an infinite more can be computed.
This is an elliptic curve, and the first rational point corresponds to {m,n} = {4,1} from which we can calculate the next, {m,n} = {17, 52}, and so on for an infinite number of points, though some of these may yield negative p or q. Source: Andrew Bremner, A Diophantine System.
2. Martin triples
A Martin triple {a,b,c} is a soln to the system,
a + b + c = x2 (eq.1)
a2 + b2 + c2 = y2 (eq.2)
a3 + b3 + c3 = z3 (eq.3)
The smallest in positive integers is M1 = {1498, 2461, 7490}, so,
1498 + 2461 + 7490 = 1072
14982 + 24612 + 74902 = 80252
14983 + 24613 + 74903 = 75973
There are also an infinite number of distinct Martin triples unscaled by a square factor, and this can be proved via a polynomial identity or an elliptic curve. Originally, A. Martin solved only eq.2 and eq.3, and gave a polynomial identity of high degree (see Form 12 ), but any soln can be made valid for eq.1 as well. Each term is simply multiplied with a scaling factor m which is the square-free part of the sum a+b+c = mn2. For ex, the above was derived from,
142 + 232 + 702 = 752
143 + 233 + 703 = 713
and since a+b+c = 14+23+70 = 107, this yields M1 after multiplying all terms by m = 107. Other distinct solns for eq.2 and eq.3 are {3, 34, 114}, {18, 349, 426}, and so on. Elliptic curves can also solve both eq.2 and eq.3, such as k = 2 of the polynomial,
Pk := (-8v2+36v+14)k + (9v2-25v+23)k + (-6v2-8v+70)k
which uses the first soln (in blue). P3 is already a perfect cube for any v, but P2 is a quartic polynomial in v which must be made a square, just like the case for Saul doublets, hence can be treated as an elliptic curve. An initial point is v = 68/75 which yields, after removing unwanted factors, the new soln,
{a,b,c} = {6213m, 32194m, 46458m}
where m = 6213 + 32194 + 46458 = 84865. From this initial point v, an infinite more can then be computed. Any Martin triple, after removing common factors, can be used to create a distinct elliptic curve similar to the one given.
Q. Are there Martin triples with GCD(a,b,c) = 1? (If we go into quadruples, then the answer is yes.)
3. Martin quadruples:
Martin quadruples {a,b,c,d} solve in non-zero integers,
a + b + c + d = x2
a2 + b2 + c2 + d2 = y2
a3 + b3 + c3 + d3 = z3
The smallest in positive integers has GCD(a,b,c,d) = 1:
m4-2m3n+2m2n2+2mn3+n4 = v2
Back to Index of Updates Back to April updates Go to July updates
(Update, 6/29/10): The eqns,
33 + 43 + 53 = 63
113 + 123 + 133 + 143 = 203
are relatively well known. Not so well known,
313 + 333 + 353 + 373 + 393 + 413 = 663
where the cubes are in arithmetic progression, differing by 2. The problem can be generalized as: “Given the sum of an n number of cubes in arithmetic progression with common difference d and initial cube a3, then for what d is this sum also a cube?” As an equation, this is,
F(d,a,n) := a3 + (a+d)3 + (a+2d)3 + …. + (a+dn-d)3 = y3 (eq.1)
Or equivalently,
F(d,a,n) := (n/4)(2a-d+dn)(2a2-2ad+2adn-d2n+d2n2) = y3 (eq.2)
For the case d = 1, there is a parametric soln, given by Dave Rusin as,
{d,a,n} = {1, (v4-3v3-2v2+4)/6, v3}
which is integral for v ≠ 3m. Note that n is a cube, a fact which will be relevant later. However, if {d,a,n} are positive integers > 1 and co-prime {d,a}, then for d < 100, a < 1000, n < 100, I was able to find several {d,a,n}. (Update, 6/30/10): Seiji Tomita extended the range to 1 < {d,a,n} < 1000. Out of the nearly 1 billion (109) possible combinations, there were only 20 primitive solns,
{2, 31, 6} {13, 28, 8} {107, 149, 3}
{3, 290, 216} {13, 230, 5} {157, 553, 7}
{5, 994, 344} {37, 15, 10} {158, 737, 14}
{7, 194, 153} {39, 824, 27} {258, 97, 8}
{11, 56, 49} {71, 435, 6} {277, 960, 16}
{11, 206, 25} {71, 816, 49} {808, 317, 3}
{11, 906, 36} {82, 541, 16}
Note that a disproportionate number of n are either squares or cubes. This phenomenon was also noted by Kevin Brown in the case d = 1. Why this is so does not seem to be known. Q. Can anyone find any other positive integers {d,a,n} > 1, and co-prime {d,a} that solves eq. 2? (Update, 2/11/11): For 4th powers in arithmetic progression d whose sum is equal to a kth power, the only known small primitive solution is d = 2 and 64 terms, given by
294 + 314 + 334 + … + 1554 = 961042
However, since the problem can be treated as an elliptic curve, there are, in fact, an infinite number of primitive solutions with 64 terms, though the initial term and d grow rapidly.
(Update, 6/27/10): Kindly check out the non-math Youtube clip at the end of this page.
(Update, 6/26/10): Some squares can be identically expressed as the sum of both n = 3 and 4 squares,
1. Form (a2+b2+c2+d2)2
n = 3: (a2+b2-c2-d2)2 + (2ad-2bc)2 + (2ac+2bd)2 = (a2+b2+c2+d2)2
n = 4: (-a2+b2+c2+d2)2 + (2ab)2 + (2ac)2 + (2ad)2 = (a2+b2+c2+d2)2
by Lebesgue and Euler. Since the form is symmetric, by judicious exchange of pairs, {b,c}, {b,d}, and {c,d}, then it can be expressed as the sum of 3 squares in 3 ways. Similarly, by choosing the negative variable from {a,b,c,d}, and adjustment of the other terms, it is the sum of 4 squares in 4 ways.
2. Form (a2+b2+2c2)2
n = 3: (2(a+b)c)2 + (2(a-b)c)2 + (a2+b2-2c2)2 = (a2+b2+2c2)2
n = 4: (a2-c2)2 + ((a+c)(b+c))2 + ((a+c)(b-c))2 + (2ac-b2-c2)2 = (a2+b2+2c2)2
both by Catalan.
3. Form (a2+b2+c2+ab+ac+bc)2
n = 3: ((a+b)(b+c))2 + ((a+c)(b+c))2 + (a2+ab+ac-bc)2 = (a2+b2+c2+ab+ac+bc)2
n = 4: a2(a+b+c)2 + b2(a+b+c)2 + c2(a+b+c)2 + (ab+ac+bc)2 = (a2+b2+c2+ab+ac+bc)2
by Catalan and J. Neuberg. Since this form is also symmetric, by cycling {a,b,c}, it is expressible as the sum of 3 squares in 3 ways.
4. Form ((a2+b2)(c2+d2))2
n = 3: ((a2+b2)(c2-d2))2 + (2cd(a2-b2))2 + (4abcd)2 = ((a2+b2)(c2+d2))2
n = 4: ((a2-b2)(c2-d2))2 + (2ab(c2-d2))2 + (2cd(a2-b2))2 + (4abcd)2 = ((a2+b2)(c2+d2))2
by Seiji Tomita. (Replacing d with b gives an old identity by Johann Euler, the son of Leonard Euler).
Q: Any other simple algebraic form that can be identically expressed both as 3 and 4 squares?
(Update, 6/23/10): 1. Saul doublets
One may observe that,
184 + 345 = 232
1842 + 3452 = 3912
1843 + 3453 = 68772
A Saul doublet {a,b} is a solution to the system of eqns,
a + b = x2
a2 + b2 = y2
a3 + b3 = z2
and named after J. Saul who first cited an example in the Gentleman’s Diary of 1795. There are an infinite number, and Bremner (1985) gave a method to completely solve for them,
Fk := (p(p+q))k + ((p+q)q)k
where,
F1 = (p+q)2
F2 = (p+q)2(p2+q2)
F3 = (p+q)4(p2-pq+q2)
hence it suffices to find,
p2+q2 = u2
p2-pq+q2 = v2
Two quadratic polynomials to be made squares define an elliptic curve, and Bremner gave an explicit recurrence relation to find all its rational points. Explicitly, one can use the formula for Pythagorean triples {p,q} = {m2-n2 , 2mn} on the first condition, and the second becomes a quartic polynomial to be made a square,
You can email the author at tpiezas@gmail.com.